A professor is interested in comparing the characteristics of students who do and do not vote in elections. For a random sample of 114 students who claimed to have voted in the last presidential election, she found a mean CPA of 2.71 and a standard deviation of 0.64. For an independent sample of 123 students who did not vote, the mean CPA was 2.79 and the standard deviation was 0.56. Test at the 5% level of significance if the mean CPA of all students who vote is lower than the CPA of students who do not vote.

The decision rule is

Fail to reject the null hypothesis

The conclusion is  

There is no sufficient evidence to conclude that the mean CPA of all students who vote is lower than the CPA of students who do not vote

Step-by-step explanation:

From the question we are told that

       The first sample size  is  

       The sample mean is  

       The standard deviation is  

      The second sample size is  

      The sample mean is  

      The level of significance is  

      The standard deviation is  

The null hypothesis is  

The alternative hypothesis is  

Generally the test hypothesis is mathematically represented as

=>

=>

Generally given that variance are not equal (the standard deviation  squared of both population are not equal ) the degree of freedom  is mathematically represented as

=>  

=>  

Generally from the t distribution table the the probability of  (t < -1.01) at a degree of freedom of is  

Generally the value obtained we see that   so

The decision rule is

Fail to reject the null hypothesis

The conclusion is  

There is no sufficient evidence to conclude that the mean CPA of all students who vote is lower than the CPA of students who do not vote

answ9x x 3=

step-by-step explanation:

1. d.90

2. e. none of these (80)

3. a. 0

4. b. 6

5. e. none of these (10)

6. a. 5